Let $P, M, N$ be $A$-modules over a field $K$. If we know that $h:P\to M$ is surjective, $g:N\to P$ is a A-homomorphism such that $hg$ is surjective, can we have $\operatorname{im} g + \ker h = P$? I think that maybe there is some $x \in P$, such that $h(x)\neq 0$ and $x \not\in \operatorname{im} g$. Thank you very much.
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$\begingroup$ I'm not sure what you mean by an $A$-module over a field $K$. But regardless, of $h\circ g$ is surjective, then you know that $g$ needs to map $N$ to submodule of $P$ isomorphic to $M$, i.e. $P/\ker h$. Hopefully that answers your question well enough? $\endgroup$– Ian ColeyMar 20, 2013 at 6:57
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It is true that there can be an $x \in P$ such that $x \notin \ker h$ and also $x \notin \operatorname{im}g$. For example consider the canonical split sequence $M \to M \oplus M \to M$ where $h(m, n) = n$ and $g(m) = (m, 0)$. Then any $m \neq 0$ gives an element $(0, m)$ which is not in the kernel of $h$ or the image of $g$.
But it is true that $P = \ker h + \operatorname{im}g$. If $p \in P$ then $h(p) \in M$. As $hg$ is surjective take $n \in N$ such that $hg(n) = h(p)$. Then $p - g(n) \in \ker h$, $g(m) \in \operatorname{im}g$, and $(p - g(n)) + g(n) = p$.
